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srebmuN desreveR (Posted on 2010-11-18) Difficulty: 3 of 5
Let us read a number k from right to left and call the result RVR(k); e.g. RVR(1)=1 , RVR(45)=54 , RVR(1200000000)=21.

1.Find the smallest positive integer k for which RVR(k) > (k^2 mod 100000).
2. Find the sum of the 10000 RVRs i.e. for all integers from 1 to 10000 inclusive.

Based upon problems from projectkhayyam.blogspot.com

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Q-1 ? | Comment 1 of 4

Q-1. 317 (smallest positive integer k for which RVR(k) > k**2 mod 100000 ) 317**2 = 100489 = 489 (mod 100000) RVR(317) = 713 713 > 489

Q-2. Sum of RVR(1) thru RVR (10000) = 499950001

Edited on November 18, 2010, 9:33 pm
  Posted by ed bottemiller on 2010-11-18 21:08:11

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